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1、Corresponding Solutions for Chemical Reaction EngineeringCHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING.1CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS.3CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA.7CHAPTER 4 INTRODUCTION TO REACTOR DESIGN.19CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR.22CHAPTER
2、 6 DESIGN FOR SINGLE REACTIONS.26CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR.32CHAPTER 11 BASICS OF NON-IDEAL FLOW. 34CHAPTER 18 SOLID CATALYZED REACTIONS.43Chapter 1 Overview of Chemical Reaction Engineering1.1 Municipal waste water treatment plant. Consider a municipal water treatment plant for
3、asmall community (Fig.Pl.l). Waste water, 32000 m3/day, flows through the treatmentplant with a mean residence time of 8 hr, air is bubbled through the tanks, and microbes inthe tank attack and break down the organic material(organic waste) +O2 ” 经 & CO2 + H2OA typical entering feed has a BOD (biolo
4、gical oxygen demand) of 200 mg 02/liter, whilethe effluent has a megligible BOD. Find the rate of reaction, or decrease in BOD in thetreatment tanks._ Clean water32,000 m3/dayWaste water Waste water32,000 m3/day Treatment plant200 mg O2needed/literMean residencetime t =8 hrZero O2neededFigure Pl.lSo
5、lution:A Vdt-32000x - ( - x day x (200 - 0) 100/32413 day L 1 OOOrn m mol32000x lja j -dayday 3 3= 18.75m。 / / ( 机3 day) = 2.17xlO 4mol/(m3 s)1.2 Coal burning electrical power station. Large central power stations (about 1000 MWelectrical) using fluiding bed combustors may be built some day (see Fig
6、.P1.2). Thesegiants would be fed 240 tons of coal/hr (90% C, 10%H2) , 50% of which would burnwithin the battery of primary fluidized beds, the other 50% elsewhere in the system. Onesuggested design would use a battery of 10 fluidized beds, each 20 m long, 4 m wide, andcontaining solids to a depth of
7、 1 m. Find the rate of reaction within the beds, based on theoxygen used.Solution:Figure PL2V = (20x4xl)xl0 = 800m3以=-240 x 103 也 x 0.5 x 0.9 kgC =-108 xl03 = -9000m)lc/(bed - hr)Az hr kgcoal hr = =4等x = 2 5 / 画处)3=9000x1 + = l2Q0Qmol/(bed - hr)dt 41 dO2 _ 1.5 xlO4 mol /(bed -hr)V dt4.17mol /(m3 s)8
8、00Chapter 2 Kinetics of Homogeneous Reactions2.1 A reaction has the stoichiometric equation A + B =2R . What is the order of reaction?Solution: Because we dont know whether it is an elementary reaction or not, we cant tellthe index of the reaction.2.2 Given the reaction 2N O2 + 1/2 O2 = N 2O5 , what
9、 is the relation between the rates offormation and disappearance of the three reaction components?Solution: - rNOi = -4& = 2%以2.3 A reaction with stoichiometric equation 0.5 A + B = R +0.5 S has the following rateexpression-FA = 2 C0.5 ACBWhat is the rate expression for this reaction if the stoichio
10、metric equation is written asA + 2B = 2R + SSolution: N o change. The stoichiometric equation cant effect the rate equation, so itdoesnt change.2.4 For the enzyme-substrate reaction of Example 2, the rate of disappearance of substrate isgiven by-fAWhat are the units of the two constants?Solution:rmo
11、l JkA E()一以 一 记7 - 6 + g 6 = CA = mol / inmol moll in 1一 一 , , in s ( mol / m3) s2.5 For the complex reaction with stoichiometry A + 3B - 2R + S and with second-orderrate expression-rA = kifABare the reaction rates related as follows: FB= FR? If the rates are not so related,then how are they related
12、? Please account for the sings , + or -.Solution: -rA - - - rB12.6 A certain reaction has a rate given by-FA = 0.005 C2 A , mol/cm3- minIf the concentration is to be expressed in mol/liter and time in hours, what would bethe value and units of the rate constant?Solution:molL-hr(-4)Xmolcm3 - minL- hr
13、mol F ) = 1()4 x 6 q =6 X1 ()4 x 0.005C； = 300C；cm min mol molLmolmolcm3-G=103CA.( 4 ) = 300C； =300x(10-3)2 =3xl()Tc；:.k =3xl() y2.7 For a gas reaction at 400 K the rate is reported as-= 3.66 p2 A, atm/hr(a) What are the units of the rate constant?(b) What is the value of the rate constant for this
14、reaction if the rate equation isexpressed as-FA = - -= k C2 A , mol/m3-sV dtSolution:(a) The unit of the rate constant is /atm- hr(b) F1 dNAV dtBecause its a gas reaction occuring at the fined terperatuse, so V=constant, andT=constant, so the equation can be reduced to( 一 丝 ) = 必2 =迹。 RT)2A VRT dt R
15、T dt RT A RT 弓= (3.66RT)C； =kC：So we can get that the value ofk = 3.66RT = 3.66 x 0.08205 x 400=120.12.9 The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol.How much faster the decomposition at 650 than at 500?Solution: 2= k = ( J_L) = 3 0 0 - _ _1_ ) = 7 586rx k R% T2 8.3
16、14 0/(10 3 加O/ .K)1173K 923K，.A = 1970.7r2.11 In the mid-nineteenth century the entomologist Henri Fabre noted that French ants(garden variety) busily bustled about their business on hot days but were rather sluggish oncool days. Checking his results with Oregon ants, I findWhat activation energy re
17、presents this change in bustliness?Running speed, m/hr150160230295370Temperature, 1316222428Solution:E Er - kQe RT f(concentratiori)let /concentration) = cons tan t - -ake RT = k e RTr r. 1 E/. LnrA = L nk-T RCr1Suppose y - LnrA,x = ., E .so slope - , intercept = Lnk/ ( ? )150160230295370LnrA-3.1780
18、-3.1135-2.7506-2.5017-2.2752T/1316222428-x lO-3T3.49473.45843.38813.36533.32061/T-y = -5147.9 x+ 15.686EAlso slope - - - - - 5147.9K, intercept = Lnk - 15.686 , = 5147.9K x8.3145J /(mol K) = 42.80V/ molChapter 3 Interpretation of Batch Reactor Data3.1 If -FA 二- (dCA/dt) =0.2 mol/liter. sec when CA =
19、 1 mol/liter, what is the rate ofreaction when CA= 10 mol/liter?N ote: the order of reaction is not known.Solution: Infonnation is not enough, so we cant answer this kind of question.3.2 Liquid a sedomposes by first-order kinetics, and in a batch reactor 50% of A isconverted in a 5-minute run. How m
20、uch longer would it take to reach 75% conversion?Solution: Because the decomposition of A is a lst-order reaction, so we can express the rateequation as:-rA= 2cWe know that for lst-order reaction, Ln - kt、gC CL n 3 = kt、 , L n - = kt2G. =0.5Q, CA2 =0.25CAOI C Cl 1So t2- t1=-(Ln - Ln = -(L n4- Ln2) =
21、 - Ln2 equ(l)k CA2 CA I k k1 C 1tx= = Ln2 = 5 min equ(2)k CA kSo t2- t =t =5 min3.3 Repeat the previous problem fbr second-order kinetics.Solution: We know that for 2nd-order reaction, - - - = kt,CA CAOSo we have two equations as follow:1 1 _ 2 11g 0=kt】= Z5min , equ(l)10 4 21 4 1 1=-=3(-)= 3k = kt2
22、, equ(2)L 4。L Ao L Ao AoSo t2 = 3。= 15 min , t2tx = 10 min3.4 A 10-minute experimental run shows that 75% of liquid reactant is converted to productby a - -order rate. What would be the fraction converted in a half-hour run?2Solution:Ina -orderreaction: -rA2人乌=5dtAfter integration, we can get:kt _ 厂
23、0.5 厂0.5y = CA O ，So we have two equations as follow:4 - ，C： ； ) = 0.59； = 他 = 女 ( 10 min), equ。 * 一 。 器 =幼 =-30m in),equ(2)Combining these two equations, we can get: 1.5C； ； = kt2, but this means C筹 0 , which isimpossible, so we can conclude that less than half hours, all the reactant is consumed u
24、p. Sothe fraction converted XA=1.3.5 In a hmogeneous isothermal liquid polymerization, 20% of the monomer disappears in34 minutes for initial monomer concentration of 0.04 and also for 0.8 mol/liter. What rateequation represents the disappearance of the monomer?Solution: The rate of reactant is inde
25、pendent of the initial concentration of monomers, sowe know the order of reaction is first-order,一 * kC, monomer - m onomercAnd Ln - = (34 min) k0.8C女 = 0.00657 min”- M onom er = 900657 min ) C _ ”3.6 After 8 minutes in a batch reactor, reactant (CA O = 1 mol/liter) is 80% converted; after 18minutes
26、, conversion is 90%. Find a rate equation to represent this reaction.Solution:In 1st order reaction, 41 , 1Lnk1 , 1Lnk 1-XA 1J V )-= 1.43, dissatisfied.Ln5)11In 2nd order reaction, 工10A CAO CAO 9/CAO 9 . r ,.- - = - - = , satisfied.J1 4/CAO 40.2Qo QAccording to the information, the reaction is a 2也o
27、rder reaction.3.7 nake-Eyes Magoo is a man of habit. For instance, his Friday evenings are all alike一intothe joint with his weeks salary of $ 180, steady gambling at 2up for two hours, thenhome to his family leaving $ 45 behind. Snake Eyess betting pattern is predictable. Healways bets in amounts pr
28、oportional to his cash at hand, and his losses are alsopredictable- at a rate proportional to his cash at hand. This week Snake-Eyes received a raisein salary, so he played fbr three hours, but as usual went home with $ 135. How muchwas his raise?Solution:nA o =180, nA - 13 , t = 2h、nA = 135 , V =3/
29、2, - rAaknArj nLn(上)L (T )So we obtain Ln = kt, -=-nA t tTL n180 L, n nA13 _ 1352 - 3=283.9 The first-order reversible liquid reactionA R , CA O = 0.5 mol/liter, CRO=Otakes place in a batch reactor. After 8 minutes, conversion of A is 33.3% while equilibriumconversion is 66.7%. Find the equation for
30、 the this reaction.Solution: Liquid reaction, which belongs to constant volume system,1st order reversible reaction, according to page56 eq. 53b, we obtainr 而 d xA ir k、t = dt = -=-Ln- 。 一 ( % +2 2 )XA 4+ & % ( % + “ 2 )XA, = 480sec = 8min, XA = 0.33, so we obtain eq(l)480 sec = 8 min - Ln- - - eq(l
31、)卜+ . 匕一( 匕+ 七) 0 33k _ CRC _ M + X/h，TA,cM = = 0 , so we obtain eq(2)2K,_ ki _ _ 3 _ 9- 二- -乙& 1 - X . _ 23k = 2k2 eq(2)Combining eq(l) and eq(2), we obtaink2 = 0.02888 min 7 = 4.8x IO_4 sec-1k = 2k2 = 0.05776 min 1 = 9.63 x 10-4 sec-1So the rate equation is - rA= - - = kCA - k2(CA o -CA)dt= 4.8xl0
32、_4sec-, CA -9.63xl0- 4s ec1-CA)3.10 Aqueous A reacts to form R (AR) and in the first minute in a batch reactor itsconcentration drops from CAO = 2.03 mol/liter to CM = 1.97 mol/liter. Find the rateequation from the reaction if the kinetics are second-order with respect to A.Solution: Its a irreversi
33、ble second-order reaction system, according to page44 eq 12, weobtain11.97 2.03 Im in,so kx =0.015-机o/minso the rate equation is q = (0.015 min )C；3.15 At room temperature sucrose is hydrolyzed by the catalytic action of the enzymesucrase as follows:sucrase 、Aucrose productsStarting with a sucrose c
34、oncentration CA O = 1.0 millimol/liter and an enzyme concentrationCEO= 0.01 millimol/liter, the following kinetic data are obtained in a batch reactor(concentrations calculated from optical rotation measurements):CA, millimol/liter 0.84 0.68 0.53 0.38 0.27 0.16 0.09 0.04 0.018 0.006 0.0025t,hr 12345
35、6789 10 11Determine whether these data can be reasonably fitted by a knietic equation of theMichaelis-Menten type, ork C C-FA = 3 where CM = Michaelis constant。 八 + CMIf the fit is reasonable, evaluate the constants ka and CM. Solve by the integral method.Solution: Solve the question by the integral
36、 method:_ r _ dA _ k?CE 0cA _ 24cA一 三T F + C w3 =容CM52 CMt1L 仁cAg o - CA 欠4CA O - CAtjtrCA jnmol/LL 鼠CAtC - C/iA o -10.841.08976.2520.681.20526.2530.531.35086.383040.381.56066.451650.271.79366.849360.162.18167.142870.092.64617.692380.043.35308.333390.0184.09109.1650100.0065.146910.0604110.00256.0065
37、11.0276cL n fcSuppose y=-, x=-, thus we obtain such straight line graphCA o - CA CA o - CA12108642(qo-d)/-y = 0.9879x + 5.0497R2 = 0.9980 1 2 3 4 5 6 7Ln(Cao/Ca)/(Cao-Ca)I r kSlope = 一 = = 0.9879、intercept= = 5.04-97k 3 Go 241 0 9879So 金 = = u，。= 0 .1 9 5 6 (./ ， ) ,k5 53970.19560.9879 xO.Ol= 19.80/
38、ir-13.18 Enzyme E catalyzes the transformation of reactant A to product R as follows:Aercyrne,R “A = 200 cAeco mol2 + g liter - minIf we introduce enzyme (CEO = 0.001 mol/liter) and reactant (CAO = 10 mol/liter)into a batch rector and let the reaction proceed, find the time needed for theconcentrati
39、on of reactant to drop to 0.025 mol/liter. N ote that the concentration ofenzyme remains unchanged during the reaction.Solution:J_ _ _ _dt_ _ 2 + g _ + 57T - JC7 - 2oo x o.ooi cA - c7+Rearranging and integrating, we obtain:Ci rO.O25 10 _r=b/ =-( 祗+5 )心= 1 0 H100 . 0 2510 L + + )+ 5(C - C J = 10 9. 7
40、 9 min100.0253.20 M . H ellin and J . C. J ungers, Bull. soc. chim. F rance, 3 86( 195 7 ) , present the data inT able P3 . 20 on thereaction of sulfuric acid with diethylsulfate in a aqueous solution at 22. 9:H 2 S O 4 + ( C2H 5 ) 2SO4 - 2 C 2 H 5 S O 4 HI nitial concentrations of H 2S O 4 and ( C2
41、H 5 ) 2SC) 4 are each 5 . 5 mol/ liter. F ind a rate equation forthis reaction.Table P3.20t, minC2H5SO4H,m ol/litert, minC2H5SO4H,m ol/liter001804.11411.181944.31481.382124.45551.632674.86752.243185.15962.753685.321273.313795.351463.764105.421623.8100(5.80)S olu tion : I ts a constant-volume system,
42、 so w e can use XA solving the problem:i) W e postulate it is a 2nd order reversible reaction system A + B 2RT he rate equation is: - rA = = kCACB -k2CdtCA ( )= CB o=5. 5mol/L, CA = CA o( l - X A)CB 一 = C4 , CR = 2 CA oXAW h e n t = g , CRe = 2 CA oXAe = 5. Smol/LC QSo = - = 0 . 5 27 3 ,A ，2x5 . 5C
43、= CB, = CA0(1 XA0 ) = 5 . 5 x( 1 0 . 5 27 3 ) = 2. 6加 。/ LA fter integrating, w e obtainL xy2 ) 1)X& = 2占eq(l)A A e X A A A eThe calculating result is presented in following Table.t,minCR, inol / LCA, mol / LXAXA,-(2X -Ln-X猾 -X,-幺 )XA,005.5000411.184.910.10730.2163-0.2275481.384.810.12540.2587-0.271
44、7551.634.6850.14820.3145-0.3299752.244.380.20360.4668-0.4881962.754.1250.250.6165-0.64271273.313.8450.30090.8140-0.84561463.763.620.34181.0089-1.04491623.813.5950.34641.0332-1.06971804.113.4450.37361.1937-1.23311944.313.3450.39181.3177-1.35912124.453.2750.40451.4150-1.45782674.863.070.44181.7730-1.8
45、1973185.152.9250.46822.1390-2.18863685.322.840.48362.4405-2.49183795.352.8250.48642.5047-2.55644105.422.790.49272.6731-2.7254005.82.60.5273X _ (2X 1)XDraw Ln - - - t plot, we obtain a straight line:XA e-XASlope = 2kl (!- - )CA o = 0.0067 ,X 0.0067= 6.794 X10- 4L /(Z?IO/-min)2(-0.5273-l)x5.5When appr
46、oach to equilibrium,K =卜=C .卜2 C tfC s eso k. = kC*A：e CBBee =6-.-7-9-4-x-1-0-*- -x-2-.-6-2- - - 1.364x10_ 4 L/(mol- mi.n) CL 5 gSo the rate equation isrA = (6.794 x 10 4 CACB -1.364 x 10*C)/w/ /(A - min)ii) We postulate it is a 1st order reversible reaction system, so the rate equation isAfter rear
47、ranging and integrating, we obtainX i ( 1 - - - ) = 7eq (2)X N e XxDraw Ln(l- - ) t plot, we obtain another straight line:A e30 100 200 300 400 500xSlope = - = -0.0068,A eSo k = -0.0068 x 0.5273 = -3.586 x 10-3 min -1,h = k- CA-e =-3-.-5-8-6-x-l-0-3-x-2-.-6- = -1t.60i7 nx 10 _3 min-CRc 5.8So the rat
48、e equation is-rA= (-3.586 x 10-3 C4+1.607 x 10-3 CR )mol/(L - nin)We find that this reaction corresponds to both a 1st and 2nd order reversible reactionsystem, by comparing eq.(l) and eq.(2), especially when XAC =0.5 , the two equations areidentical. This means these two equations would have almost
49、the same fitness of data whenthe experiment data of the reaction show that XAC =0.5.(The data that we use just have XAC=0.5273 approached to 0.5, so it causes to this.)3.24 In the presence of a homogeneous catalyst of given concentration, aqueous reactant Ais converted to product at the following ra
50、tes, and CA alone determines this rate:CA,mol/liter 1 2 4 6 7 9 12-FA, mol/liter hr 0.06 0.1 0.25 1.0 2.0 1.0 0.5We plan to run this reaction in a batch reactor at the same catelyst concentration as used ingetting the above data. Find the time needed to lower the concentration of A from CAO = 10mol/
51、liter to CM = 2 mol/liter.Solution: By using graphical integration method, we obtain that the shaped area is 50 hr.203.31 The thermal decomposition of hydrogen iodide2HI H2 + bis reported by M.Bodenstein Z.phys.chem.,29,295( 1899) as follows:T/C 508 427 393 356 283k,cm3/mob 0.1059 0.00310 0.000588 8
52、0.9x1 O -6 0.942x1 O -6Find the complete rate equation for this reaction. Use units of joules, moles, cm3, andseconds.According to Arrhenius9 Law,k 二 koe-E/R Ttransform it,-In(k) = E/R (l/T) -In(ko)Drawing the figure of the relationship between k and T as follows:From the figure, we getslope 二 E/R =
53、 7319.1 intercept = - In(ko)= -11.567E = 60851 J/mol ko = 105556 cm3/mol-sFrom the unit k we obtain the thermal decomposition is second-order reaction, so the rateexpression isT A= 105556e 6085,/RT-CA2Chapter 4 Introduction to Reactor Design4.1 Given a gaseous feed, CAO = 100, CBO = 200, A + B 1 R +
54、 S, XA = 0.8. Find XB,CA,CB.Solution: Given a gaseous feed, CA o = 100 , CBo = 200 , A + B R + S=0,find XB, CA, CBV。 . CA= CAo( l - X J = 100x0.2 = 20_ 1x100 x0.8 _2A D U.4+CB0 200CB =Cfio(l- XB) = 200 x 0.6 = 1204.2 Given a dilute aqueous feed, CAO = CBO =100, A + 2 B - R + S, CA = 20. Find XA, XB,
55、CB.Solution: Given a dilute aqueous feed, CA l - CBo - 100 ,A+2B - H + S,= 2 0 ,find X . X“ ，CKAqueous reaction system, so A= eK=0When XA= 0, V = 200When X . =1, V = 100S o以ACB o _ _ j_而z1 q = 1 .里0.8,b CA oXA 2 100x0.8- - X -= 1.6 1, which is impossible.So XB= L C-004.3 Given a gaseous feed, CAO =2
56、00, CBO =100, A +B R, CA = 50. Find XA, XB, CB.Solution: Given a gaseous feed, CA o = 200 , CBo100,A + Bf R CA = 5 0 .find XA, XB, CBX . = 1 - - C - = 1 -50- = 0.75 ,鼠 200X B = = 1.5 1, which is impossible.CB OSo CB= CB. = 1004.4 Given a gaseous feed, CAO = CBO =100, A +2B R, CB = 20. Find XA, XB, C
57、A.Solution: Given a gaseous feed, CA n + CBo = 100 , A + 2B f R , CBo = 20 .Find 乂XB，CAXB = 0 , V = 100A + 1003 = 200XB =1, V = 50A + 100R = 150150-200200 0.25, A-0.25x100-0.5xlOOXB100-20100-0.25x20= 0.842,*x 100x0.8422100X .= 0.42112C4A = CAo- = 100 x -：-Aol + AXA 1 -0 .5X0.421= 73.344.6 Given a ga
58、seous feed, To =1000 K ,兀o=5atm , CAO=1OO, CBO=2OO, A + B -5 R ,T =400 K,7i=4atm, CA =20. Find XA, XB, CB.Solution: Given a gaseous feed, To = 1000 K , = 5atm , CA o = 100 , CBo = 200A + 8 5R , T = 4 0 0 K ,乃= 4 m , CA = 20, find XA, Q .6 0 0 -3 0 0 aACBo 400 x5 300 bCA n T。 兀 1000 x 4According to e
59、q page 87,XACA o 20 n.】 厂 T 1- x 0.5CA To 兀一 100l + -C-A- -T-TV-O 1 + lx 20x0.5CA“ T。兀 100= 0.818XBbCA oXA _ _ 100 x 0.818aCHo - 2 0 0= 0.409cBot I jtoo-o-818)x2O13O1 + XA1 + 1x0.8184.7 A Commercial Popcorn Popping Popcorn Popper. We are constructing a 1-liter popcornto be operatedin steady flow. Fi
60、rst tests in this unit show that 1 liter/min of raw corn feed streamproduces 28 liter/minof mixed exit stream. Independent tests show that when raw corn pops its volume goesfrom 1 to 31.With this information determine what fraction of raw com is popped in the unit.Solution: sA =-! !- = 30, CA o = a.
61、u., CA = -CA ( =-a.u.1 Zo Zo1 -J -X. = C . Y = -= 46.5%CA +ACA I + 3 0X 28Chapter 5 Ideal Reactor for a single Reactor5.1 Consider a gas-phase reaction 2A 一 R + 2s with unknown kinetics. If a space velocityof 1/min is needed for 90% conversion of A in a plug flow reactor, find thecorresponding space
62、-time and mean residence time or holding time of fluid in the plugflow reactor.Solution: r = = 1 min ,sVarying volume system, so t cant be found.5.2 In an isothermal batch reactor 70% of a liquid reactant is converted in 13 min. Whatspace-time and space-velocity are needed to effect this conversion
63、in a plug flowreactor and in a mixed flow reactor?Solution: Liquid reaction system, so 川=0According to eq.4 on page 92, t = g 。 广 A = 心、 FQ _Q C XEq. 13, % F R = - = -， M F R cant be certain.-rA -rAEq-17, TP F R = CA o 笠， so M R = = 13 min5.4 We plan to replace our present mixed flow reactor with on
64、e having double the bolume.For the same aqueous feed (10 mol A/liter) and the same feed rate find the newconversion. The reaction are represented byA - R, -FA = kC 1.5 ASolution: Liquid reaction system, so 八 = 。心 点 -kCA o( l-XA)-5N ow we know: V = 2V, = FA o, c j =CA o, X“ =0.7So we obtain2V X； _ 2X
65、FAO 几 X； )L5 左或( ) X j5X； = 2x0.7 8.52(1-X， 5 d-0.7)5X； = 0.7945.5 An aqueous feed of A and B (4001iter/min, 100 mmol A/liter, 200 mmol B/liter) is to beconverted to product in a plug flow reactor. The kinetics of the reaction is representedbyJTinlA +B R, -rA = 200CACB 空 一liter-minFind the volume of
66、 reactor needed for 99.9% conversion of A to product.Solution: Aqueous reaction system, so = 0According to page 102 eq. 19, - = - = - j - = j -FA O AO go rA 一 以r = = CA o f , vo - 4(X) liter/ min ,rXA f dX . f0.999 dX A= = 0x400( L = 124.3L-rA - rA5.9 A specific enzyme acts as catalyst in the fermen
67、tation of reactant A. At a given enzymeconcentration in the aqueous feed stream (25 liter/min) find the volume of plug flowreactor needed for 95% conversion of reactant A (CA O =2 mol/liter ). The kinetics of thefermentation at this enzyme concentration is given byA enzyme、D - ，ITlolA -:- K , -LA -1
68、 + 0.5C4 min - literSolution: P.F.R, according to page 102 eq.l 8, aqueous reaction, c = 0V exA dXA乙J。fV = F .,厂1 + 05。4 = 125 x n_ 1 _+ x )4 J。 0.1Cx - XA A= 125(L + 0.95) = 986.4L0.055.11 Enzyme E catalyses the fermentation of substrate A (the reactant) to product R. Findthe size of mixed flow rea
69、ctor needed for 95% conversion of reactant in a feed stream(25 liter/min ) of reactant (2 mol/liter) and enzyme. The kinetics of the fermentation atthis enzyme concentration are given by-FA =0.1CA mol1 + 0.5g min literSolution: vo = 25L/tnin , CA o = 2tml / L. FA o = 50mol/rvin , XA - 0.95Constant v
70、olume system, M.F,R., so we obtain2x0.950.1x0.05x2= 199.5 min,1 + 0.5x0.05x2V - TVO = 199.5min x 25L/min = 4.9875m35.14 A stream of pure gaseous reactant A (CA O = 660 mmol/liter) enters a plug flow reactor ata flow rate of FA O = 540 mmol/min and polymerizes the as follows 八 “ mmol3A 一 R, -FA = 54-
71、 /minHow large a reactor is needed to lower the concentration of A in the exit stream to CM- 330 mmol/liter?1 , 1 CA 330-1 1- 1-Solution:屋= = XA= = J鸳= 0.754 1 3 A . CA . 2 330l + . 1 x, Cv, 3 6600-order homogeneous reaction, according to page 103 eq.20So we obtainV = CA -XA =540x= 7.5LC&, k A 545.1
72、6 Gaseous reactant A decomposes as follows:A - 3 R, -FA = (0.6min1)CAFind the conversion of A in a 50% A 一 50% inert feed (uo = 180 liter/min, CAO =300mmol/liter) to a 1 m3 mixed flow reactor.a 4 -2Solution: V = 1/W3,M.ER. = - = 12According to page 91 eq. 11,忆 F 06。 匕区1 + XA= X/ + XA) = 1000- 0 .6 (
73、l-X J - 1 8 d /m inSo we obtain XA = 0.667Chapter 6 Design for Single Reactions6.1 A liquid reactant stream (1 mol/liter) passes through two mixed flow reactors in aseries. The concentration of A in the exit of the first reactor is 0.5 mol/liter. Find theconcentration in the exit stream of the secon
74、d reactor. The reaction is second-orderwith respect to A and V2/V1 =2.Solution:V _ go gi T_ _ 匕 _ gi C,U()l - rA U02 rA2CA()=lmol/l, CAi=0.5mol/l, = P02 , -rAi=kC2 Al ,-FA2=kC2 A2 (2nd-order), 2x- CM CM - CA9So we obtain2x( 1 -0.5)/(k0.52)=(0.5-CA2) /(kCA22)CA2= 0.25 mol/16.2 Water containing a shor
75、t-lived radioactive species flows continuously through awell-mixed holdup tank. This gives time for the radioactive material to decay intoharmless waste. As it now operates, the activity of the exit stream is 1/7 of the feedstream. This is not bad, but wed like to lower it still more.One of our offi
76、ce secretaries suggests that we insert a baffle down the middle of thetank so that the holdup tank acts as two well-mixed tanks in series. Do you think thiswould help? If not, tell why; if so calculate the expected activity of the exit streamcompared to the entering stream.Solution: Ist-order reacti
77、on, constant volume system. From the information offered aboutthe first reaction,we obtain% KCM * CA。If a baffle is added,_ 21 ,匕2 _ 匕 _ A 0 一 C 4 21 4 A 2 一 A 2 2- 1 - -1 -匕。Y-=6/k% 21 % 22沁。lyC40 C 42 _ 2 1 =3/k二 一0A22Combining equation and we obtain:CA2I= O. 25 CA O ;CA22=0 . 25 CA2I= CM)16 ASo i
78、t will help, and the expected activity of the exit stream is 1/16 of the feed.6.3 An aqueous reactant stream (4 mol A/liter) passes through a mixed flow reactorfollowed by a plug flow reactor. Find the concentration at the exit of the plug flowreactor if in the mixed flow reactor CA = 1 mol/liter. T
79、he reaction is second-order withrespect to A, and the volume of the plug flow unit is three times that of the mixed flowunit.Solution: Constant volume system and 2nd-order reaction:Combining equation. and we obtain:CA尸 0.1 mol/liter6.4 Reactant A (A R, CA O=26 mol/m3) passes in steady flow through f
80、our equal-sizemixed flow reactors in series (r total =2 min). When steady state is achieved theconcentration of A is found to be 11, 5, 2, 1 mol/m3 in the four units. For this reaction,what must be r plug so as to reduce CA from CA O = 26 to CM = 1 mol/m3?Solution:_ CA O _ CA I _ A I A 2 _ A 2 _ C 4
81、3 _ C 43 - A 42 = = TmA -一-。- - -rA2 rA3 rA4CAo=26mol/liter, CA I=1 1 mol/liter, CA2=5 mol/liter, CA3 = 2mol/liter, CA4 =1 mol/literSo we abtain: 15/(-rAi) = 6/ ( -F A2) = 3 / ( -I A3 ) = l/(-rA4)We postalate the reaction rate is 1 unit when CA4 =1 mol/literSo we obtainCA, mol11 5 2 1-FA30 12 6 2l/(
82、-rA)1/30 1/12 1/6 1/2 P Jr. “ J r . , ”So we obtain TP = 2.63 min.volume batch reactor as follows:6.6 At 100 pure gaseous A reacts away with stoichiometry 2A R + S in a constantt, sec020406080100120140160PA, atm1.000.960.800.560.320.180.080.040.02What size of plug flow reactor operating at 100 and 1
83、 atm can treat 100 moles A/hr in afeed consisting of 20% inserts to obtain 95% conversion of A?Solution:FA O= 100 mol/hr = 0.0278 mol A/s, CA O = (1 atmxl01.3)/(8.314x37.3)=0.0327mol/lt, sec020406080100120140160PA, atm1.000.960.800.560.320.180.080.040.02CA,mol/10.03270.031360.02610.02850.010450.0058
84、80.002610.0013070.0006533N ow , PA O =0,8 atm, CA0 =0.0261 mol/1When the conversion of A is 90%, CA =0.001305mol/l, pA =0.04 atmSo T = r0 04 - r0 8= 140-40= 100 sSo v = w0 = r - = l 00x0.027/0.0261 = 106.4 Lg 。6.7 We wish to treat 10 liters/min of liquid feed containing Imol A/liter to 99% conversio
85、n.The stoichiometry and kinetics of the reaction are given byA R,-FA =CA mol0.2 + liter- minSuggest a good arrangement for doing this using two mixed flow reactors, and findthe size of the two units needed. Sketch the final design chosen.Solution:L 0=101/min , CA O= 1 mol/1,when ZA=0.99 CAf=0.91mol/
86、l(1) two for equal-she M.F.R.匕 _ CA O 一 Gu _ Cm - 0二% 一% rA fIA产CAI/(0.2+CAI) , -rAf= CAf/(0.2+CAf) = 0.0476,So we obtain CAI=0.121 mol/1, r i=2.333minV(=2Vi=2 r,p0 =2x2.333x10=46.65 L(2)When the area of rectangle KLMN is maximum, the volume of reactors needed its minimum.r =乜 + 畛=CAOF + 勿=2.7+1.89=
87、4.590。4 f -rA 2So =4.59x10=45.9d(-)Slope of curse = - - =0.2/( 1-ZA)2dx1 02A (-) 21-(匚 + 1)Slope of LN = - - =- - Z 川 ZzSo when slope of curse = slope of LN , ZAI=0.9,CAI =0.1 mol/L, -FAI=1/3 , l/(-rA) = 36.8 From steady-state kinetics runs in a mixed flow reactor, we obtain the following dataon the
88、 reaction A R.T , secCAO, mmol/ literCA, mmol/ liter605 0203 510 04 01110 0602020 0801120 010 0F ind the space time needed to treat a feed of CAO= 10 0 mmol/ liter to 8 0 %conversion( a) in a plug flow reactor.( b) in a mixed flow reactor.Solution: F rom the data offered, we obtain( a) in a P. F . R
89、: CAO=1OO mmol/ L , XA=0.8SoCA= 20 mmol/ L TP =area of the shaded( b) in a M . F . R.rm - =( 10 0 -20 ) / ( -rA 2( ) ) =2x80 = 160 s.6.20 Reactant A decomposes with stoichiometry A R and with rate dependent only onCA. T he following data on this aqueous decomposition are obtained in a mixed flow rea
90、ctor:T , secCAOCA1420 010 0251909029180803 017 07 029160602715 05 02414 04 01913 03 0151202012110102010 11Determine which setup, plug flow, mixed flow, or any two-reactor combination givesminimum r for 9 0 % conversion of a feed consisting of CAO = 10 0 . A lso find thisT minimum. I f a two-reactor
91、scheme is found to be optimum, give CA betweenstages and r for each stage.Solution :Draw a 一 CA curseCAO= 100, ZA=0.9, SO CA=10From the CA curse, we know that when CA ( 10,70), increases as CAincreases and that when CA (70,100),- decrease as CA increase. So when CA e (10,70),rAwe plan to use a P.F.R
92、 TP = area of the shaded region followed with a M.F.R. to treat the feedCA 10,70), rm = area of the shaded region.Chapter 10 Choosing the Right Kind of Reactor10.1 Given the two reactionsA + B R -ri = RICACBR + B S -r2 =IQCACBWhere R is the desired product and is to be maximized. Rate the four sohem
93、es shown inFig. P10.1 一either good or not so good, please, no complicated calculations, justreason it out.Solution: N umber (d) is good for the formation of intermediate -R which causes nomaximum occurring there. rR = (- )-(- ) = - k2)CACH, the same order, so R cantaffect the D.10.2 Repeat Problem 1
94、 with just one change-F2 = 1QCRC2 BSolution:-n = RICACB n=2-F2 = R2CRCB2 n2=3m n2, therefore, low CB favors the reaction of lowest order, mixed flow reactor should beused.So (a) N ot so good. (b) Good.(c) N ot so good. (d) N ot so good10.3 Repeat Problem 1 with just one change-r2 = k2c2 RC BSolution
95、:-n = RICACB m=2-F2 = k2CR2cB 112=3m 0.05 力(b) r = - = - = 2.5 minC力 10.05力Jo Jor = , so V = r - v = 2.5minx4/min = 10LvCE = U 0.05x4 = 0.2min tM /v11.2 Repeat problem P ll.l with one change: the tracer curve is now as shown in Fig.Pl 1.2.0.051Figure P11.2MSolution: Ac_t = 0.05x6 = 0.30 vSo the resu
96、lts are not consistent.11.3 A pulse input to a vessel gives the results shown in Fig. Pl 1.3.(a) Are the results consistent? (check the material balance with the experimental tracercurve.)(b) If the results are consistent, determine the amount of tracer introduced M, and the Ecurve.v = 4 cm3/sV= 60
97、cm3Figure P11.3Solution:1 9(a) Ac_, = -/zx(25-16) = - /I,7口 闻 J ：子19 25Lcdt+LcdtQ5 cm a x(t - 25)/力- = 20sec(t-25)dt19616 + 19 C,一x3x2 2,m a x2max19 + 25 ,+ -x6xx(25-16)cmax2- 2 - = 20.5 secVv竺 = 1 5 s e c 4So the results are not consistent.11.4 A step experiment is made on a reactor. The results ar
98、e shown in Fig. P11.4.(a) Is the material balance consistent with the tracer curve?(b) If so, determine the vessel volume V, i ,the F curve and the E curve.u = 4 liters/minm = 0.5 mol/minFigure P11.4Solution:(a) If Cniaxm 0.5 - 1 + 3一= 4 = 0.125moiI L , A = max =-c- -C.i2max2C/ J n a xso t = 2 min(b
99、) F =匕C = -比 Cmax0, re 0,1)then F curse is F =10 yr) and the data of Fig.P11.5 are obtained.(a) How many units of this tracer were introduced into the river?(b) What is the volume of Columbia River waters between Bonneville Dam and the pointof introduction of tracer?E、A】=。P. 2PB+0s-cn-DSolution:(a)
100、x IO * 105 = 5.25x M 二. 助, = ，2 m vr根3so we obtain M = 5.25 x 10 - - day x 6000 = 27216 units of radioisotopems(b)- _ r tCdt,- J。CdtIO-635-20/I 258 2。5t35-125(z -125) Jr；x (125-2 0 ”10一63.15x10-352.5x10-6二60d av = i W = 60day X 6000 = 3.11x 10/311.6 A pipeline (10 cm I.D., 19.1 m long) simultaneousl
101、y transports gas and liquid fromhere to there. The volumetric flow rate of gas and liquid are 60 000 cm3/s and 300 cm3/s,respectively. Pulse tracer tests on the fluids flowing through the pipe give results as shown inFig.P11.6. What fraction of the pipe is occupied by gas and what fraction by liquid
102、?Figure P11.6Solution:鼠= 2 s, tj = 1005, so 匕=乙=6000x2 = 1.2xl05cm3Vj = v = 300x100 = 3xl04cm3total = -4 2A = -4X102 X19.1X102 =1.5xl05cm3i 2x 105So we obtain %G = - 100% = 80%, %L = 20%1.5xl05A liquid macrofluid reacts according to A R as it flows through a vessel. Find theconversion of A for the f
103、low patterns of Fig.P11.7 to Pl 1.11 and kinetics as shown.11.7 CAO = 1 mol/liter, -FA = kC0.5 A, k = 2 mol05/liter0 5-minE, min”Figure 11.7So we obtain 削 =( 3) . E力=j ： (1T ) 2 g 力=：/. X. =1 = 0.8334 611.8 CAO = 2moi/liter, YA = kC2 A, k = 2 liter / mol-minFigure 11.8Solution:According to page274 e
104、q.15,g - 1go 1 + C /AE_f= l9so m ax=3m in-14r+ 1J 0,0.5E = 0,r G (0.5,+oo)So普 =仔 ) E . dt = f +1油= 0.5,XA =1-0.5 = 0.511.9 CAO = 6mol/liter, -FA = k, k = 3 mol/1 iter-minE, min-111.11 CA O = O. lmol/ liter, -F A = k, k =0 . 0 3 mol/ liter minSolution:s o11.10 CA O = 4 mol/ liter, -F A =Solution:Sofi
105、 kt c .Q 1-2minAE_t = 1, so = 4 min 、XA = 1 - 4 = 0 . 7 5A 4Figure 11.9Solution: For the zero-order reaction,So(XQelementkt fl-O.3r,z 3.3minelement -0.3/,/ 3.3min- pcoXA = (XQ Edt (liquid macrofluid reacts)*0 elementFrom Fig 11.11, we knowEQ) =0,0 r 4 min0.1(， -4),4 t 6min0.025(14-0,6 r 14min- r 14=
106、 1 1 E(t)dt = 1Thus, the conversion of A is 100%.11.12 11.14 Hydrogen sulfide is removed form coal gas by contact with a moving bed ofiron oxideparticles which convert to the sulfide as follows:Fe2Os FeSIn our reactor the fraction of oxide converted in any particle is any particle isdetermined by it
107、s residence time t and the time needed for complete conversion of theparticle r and this is given byt a1 - X = (1- - y when t 1 hrFind the conversion of iron oxide to sulfide if the RTD of solides in thecontactor is approximated by the curve of Fig.Pl 1.12,P 11.13,P 11.14.11.12hr-1Figure 11.12Soluti
108、on:F rom F ig. 11. 12, we know1, 0 r I hrWhentvlhr, 1 -( XALS“ = (1T)3So 匹=1) e：E dt = 1 x 1 - ( 1 - r) 3 力=0 . 7 511.130.5Figure 11.13Solution: F rom F ig. 11. 13 , we knowEQ) 二ooj = 0 . 5 / ir0 J w 0 5 hrW h e n t 1 hr, 1 - ( X A )etemeB, = ( 1 - 03also (X Q c , = 1- ( 17 ) 3So意=（ x A %即）力=- （ i -
109、 厅 a- o. 5 ）力0 . 87 5T hus, the conversion of A is 87 . 5 % .11.14i, hrFigure 11.14F rom F ig. 11. 13 , we knowO, ZO. 5 / zrE( t) = 1, 0 . 5 t 1. 5 / zrWhentlhr, ( XA)elemenl = 1So = ( XA力=f 1 一( 1 -) 3 1力 + Jlxldt = 0 . 984 4T hus, the conversion of A is 98. 4 4 % .Chapter 18 Solid Catalyzed Reacti
110、ons18.22 In the absence of pore diffusion resistance a particular first-order gasphase reactionproceeds as reported below.-吸= 10-6 mol/cm3cat-sat CA = 10-5 mol/cm3, at 1 atm and 400what size of spherical catalyst pellets ( 韵6 - 10-3 cm/cm cat- s) would ensure thatpore resistance effects do not intru
111、de to slow the rate of reaction?Solution:，The rate expression of first-order reaction is = kmCAAnd with -吸, CA replaced by numerical values, we obtainkIO-610= 0.1To judge what type of the pore resistance, we need to calculate the Thiele modulusWhen MT, the pore resistance effects could be neglected.
112、3Thus, -dp 0.4, also dp 24mmSo the diameter of spherical catalyst pellets should be less than 2.4mm.18.24 The first-order decomposition of A is run in an experimental mixed flow reactor.Find the role played by pore diffusion in these runs; in effect determine whether the runs weremade under diffusio
113、n-free, strong resistance, or intermediate conditions.dp W CAO U XA4 1 300 60 0.8 AR8 3 100 160 0.6Solution: - rA ou, =w_ _ . , 300x60x0.8 t _For the first run, - rA outX =- = 1 4 4 CFor the second run,-rA o a f1 0 0:1 6 0:0.6 . r e- = 320314400 ， - = 4.53200While& _ dPa _ 2Thus, the runs were running under intermediate pore resistance.
